Perfect Squares
Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...
) which sum to n.
For example, given n = 12
, return 3
because 12 = 4 + 4 + 4
; given n = 13
, return 2
because 13 = 4 + 9
.
解題思路
如果
可以利用動態規劃(Dynamic Programming)求解該問題。令
代碼如下:
class Solution {
public:
int numSquares(int n) {
int dp[n+1];
// 將所有非平方數的結果置最大
fill(dp, dp + n + 1, INT_MAX);
for (int i = 0; i * i <= n; ++i) {
// 將所有平方數的結果置1
dp[i * i] = 1;
}
for (int i = 0; i <= n; ++i) {
for (int j = 0; i + j * j <= n; ++j) {
// 注意 a + b*b 本身就是平方數
dp[i + j * j] = min(dp[i] + 1, dp[i + j * j]);
}
}
return dp[n];
}
};